Vectors in components, magnitudes and bearings · Vectors

In General Maths you used column vectors, found their lengths, and added them head to tail (see vectors). Further Maths writes the same vectors with the unit vectors $\mathbf i$ (one step east) and $\mathbf j$ (one step north): $\begin{pmatrix} 4 \\ 3 \end{pmatrix} = 4\mathbf i + 3\mathbf j$ . It then asks for lengths, directions and bearings.

Magnitude|xi + yj| = √(x² + y²)

Unit vectors and vectors of a given length

A unit vector has length 1. To get one in the direction of $\mathbf a$ , divide $\mathbf a$ by its length:

A unit vectorâ = a ÷ |a| points the same way, with length 1

A vector of length $k$ in the direction of $\mathbf a$ is then $k\,\hat{\mathbf a} = \dfrac{k}{|\mathbf a|}\mathbf a$ .

Worked example · WAEC 2011

WAEC 2011 · Paper 2 · Q7

Given that $\mathbf n = \begin{pmatrix} -12 \\ 5 \end{pmatrix}$ and $\mathbf s = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$ , find the vector $\mathbf q$ such that $|\mathbf q| = 35$ and $\mathbf q$ is in the direction of $(\mathbf n + 5\mathbf s)$ .

The direction
- ${\mathbf n + 5\mathbf s = \begin{pmatrix} -12 + 5 \\ 5 - 5 \end{pmatrix} = \begin{pmatrix} -7 \\ 0 \end{pmatrix}}$ .
Think first. Work out n + 5s.
The unit vector
- ${|\mathbf n + 5\mathbf s| = 7}$ , so the unit vector is ${\begin{pmatrix} -1 \\ 0 \end{pmatrix}}$ .
Think first. Its length is 7. Divide by it.
Length 35
- ${\mathbf q = 35\begin{pmatrix} -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -35 \\ 0 \end{pmatrix}}$ .

More: magnitudes and unit vectors

Magnitude and bearing

A vector can also be given as a length and a bearing, such as $(8\text{ N}, 135^\circ)$ . The bearing is measured clockwise from north, so the east part uses sine and the north part uses cosine:

From a bearing to components(r, θ) = r sin θ i + r cos θ j

Going back, the length is $\sqrt{x^2 + y^2}$ , and the bearing comes from $\tan^{-1}$ of east over north, adjusted for the quadrant. Draw a quick sketch to see which quadrant the vector is in.

Worked example · WAEC 2016

WAEC 2016 · Paper 2 · Q15 (a)

Given that $\mathbf m = 6\mathbf i + 8\mathbf j$ and $\mathbf n = -8\mathbf i + \frac73\mathbf j$ , find, correct to two decimal places, the magnitudes and directions (bearings) of $\mathbf m$ and $\mathbf n$ .

The magnitude and bearing of m
- ${|\mathbf m| = \sqrt{36 + 64} = 10}$ .
- East 6, north 8: the angle from north is ${\tan^{-1}\frac68 = 36.87^\circ}$ , so the bearing is ${036.87^\circ}$ .
Think first. m = 6i + 8j points north-east. What is the angle from north?
The magnitude of n
- ${|\mathbf n| = \sqrt{64 + \frac{49}{9}} = \sqrt{\frac{625}{9}} = \frac{25}{3} \approx 8.33}$ .
Think first. n = −8i + 7/3 j.
The bearing of n
- It points north-west. The angle from north towards west is ${\tan^{-1}\frac{8}{7/3} = \tan^{-1}\frac{24}{7} = 73.74^\circ}$ .
- So the bearing is ${360^\circ - 73.74^\circ = 286.26^\circ}$ .
Think first. West 8, north 7/3: which quadrant?

More: magnitudes and bearings

Resultants of several forces

To add forces given by bearings, change each to components, add the $\mathbf i$ parts and the $\mathbf j$ parts, then change the total back. The force that keeps a body in equilibrium (the equilibrant) is the resultant reversed.

Forces by size and bearingSet each force

9.99i − 0.55jresultant, i and j10 N, 93.1°magnitude and bearing

F₁ (N) = 6F₁ bearing = 040°F₂ (N) = 8F₂ bearing = 130°

F₁ = 6 sin 40° i + 6 cos 40° j = 3.86i + 4.6j; F₂ = 6.13i − 5.14j. Add the components, then the resultant has size √(9.99² + (−0.55)²) = 10 N.

Worked example · WAEC 2017

WAEC 2017 · Paper 2 · Q8

Forces $F_1(18\text{ N}, 330^\circ)$ , $F_2(10\text{ N}, 090^\circ)$ and $F_3(25\text{ N}, 180^\circ)$ act on a body at rest. Find, correct to one decimal place, the magnitude and direction of the resultant force.

Components
- ${F_1}$ : ${18\sin 330^\circ = -9}$ and ${18\cos 330^\circ = 15.588}$ .
- ${F_2}$ : ${10\sin 90^\circ = 10}$ and ${10\cos 90^\circ = 0}$ .
- ${F_3}$ : ${25\sin 180^\circ = 0}$ and ${25\cos 180^\circ = -25}$ .
Think first. East part r sin θ, north part r cos θ, for each force.
Add them
- East: ${-9 + 10 + 0 = 1}$ .
- North: ${15.588 + 0 - 25 = -9.412}$ .
Magnitude and bearing
- ${|\mathbf R| = \sqrt{1^2 + 9.412^2} = \sqrt{89.59} \approx 9.5}$ N.
- It points south-east: the angle east of south is ${\tan^{-1}\frac{1}{9.412} = 6.07^\circ}$ .
- So the bearing is ${180^\circ - 6.07^\circ \approx 173.9^\circ}$ .
Think first. 1 east and 9.412 south: which quadrant?

More: resultants of forces

Unknown scalars

When one vector is written as $m\mathbf a + n\mathbf b$ , match the $\mathbf i$ parts and the $\mathbf j$ parts. That gives two equations in $m$ and $n$ .

More: unknown scalars

Your turn

WAEC 2022 · Paper 2 · Q15

Vectors

The vectors $6\mathbf i + 8\mathbf j$ and $8\mathbf i - 6\mathbf j$ are parallel to $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ respectively. If the magnitudes of $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ are 80 units and 120 units respectively, express:

(a)
$\overrightarrow{OP}$ and $\overrightarrow{OQ}$ in terms of $\mathbf i$ and $\mathbf j$ ;
Show the answer
$\overrightarrow{OP} = 48\mathbf i + 64\mathbf j$ , $\overrightarrow{OQ} = 96\mathbf i - 72\mathbf j$
(b)
$|\overrightarrow{PQ}|$ in the form $c\sqrt k$ , where $c$ and $k$ are constants.

Worked solution (try it first)

(a)

$|6\mathbf i + 8\mathbf j| = 10$ , so $\overrightarrow{OP} = \frac{80}{10}(6\mathbf i + 8\mathbf j)$
$= 48\mathbf i + 64\mathbf j$ .
$|8\mathbf i - 6\mathbf j| = 10$ , so $\overrightarrow{OQ} = \frac{120}{10}(8\mathbf i - 6\mathbf j)$
$= 96\mathbf i - 72\mathbf j$ .

(b)

$\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}$
$= 48\mathbf i - 136\mathbf j$ .
$|\overrightarrow{PQ}| = \sqrt{2304 + 18\,496}$
$= \sqrt{20\,800}$ .
$20\,800 = 1600 \times 13$ , so $|\overrightarrow{PQ}| = 40\sqrt{13}$ .

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